Zn + H2SO4 → ZnSO4 + H2
65x 0,35 mol
ZnO + H2SO4 → ZnSO4 + H2O
81x
\(n_{H_2}=\dfrac{7,84}{22,4}=0,35\) (mol)
Theo PT : \(n_{H_2}=n_{Zn}=0,35\) (mol)
→ mZn=0,35 . 65 = 22,75 (g)
→%Zn = \(\dfrac{22,75}{25,95}.100\%\approx87,7\%\)
→%ZnO = 100% - 87,7% = 12,3 %